# Potential Energy in General Relativity

I often hear about how general relativity is very complicated because of all forms of energy are considered, including gravitation's own gravitational binding energy. I have two questions:

1. In general relativity, objects following the motion of gravitation should simply be travelling by geodesics. In such 'free fall', why would there be any 'binding energy'?
2. From Einstein's field equations, $$R_{\mu \nu} - {1 \over 2}g_{\mu \nu}\,R = {8 \pi G \over c^4} T_{\mu \nu},$$ isn't the curvature only coupled to the energy-momentum tensor? As far as I understand, potential energy is not included inside the energy-momentum tensor.
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 Are you asking about gravitational self energy? If so a quick Google will find you lots of stuff. Gravitational waves carry energy, so by extension gravitational fields do as well. – John Rennie Nov 26 '12 at 17:39 Your first question seems unrelated to the others. It could be phrased better as "What's the potential energy experienced by a test particle following geodesic motion? Why would a geodesic experience 'binding energy'?" – Alex Nelson Nov 26 '12 at 17:45

In GTR, these constants are given by a Killing vector field, which is an infinitesimal generator of an isometry: spacetime "looks the same" in the direction of a Killing vector. Most spacetimes do not have any, but by definition, a static spacetime has a timelike Killing vector field, and can always be put in the following form: $$ds^2 = -\lambda dt^2 + d\Sigma^2,$$ where $d\Sigma^2$ is the metric for any spacelike manifold and $\lambda$ is independent of $t$. The factor $\lambda^{1/2}$ is commonly called the gravitational redshift.
For example, for the Scwarzschild spacetime in the usual Schwarzschild coordinates, $\lambda = \left(1-\frac{2GM}{c^2R}\right)$, and the following is a constant of motion representing the specific (per-mass) energy of the freefalling particle: $$e = \left(1-\frac{2GM}{c^2r}\right)\frac{dt}{d\tau}.$$ This is the natural generalization of the total mechanical energy, including also rest-mass energy; for the Schwarzschild case, the spherical symmetry allows one to build an "effective potential" quite analogous to the Newtonian case, but that approach is less useful in general.